2. please use Gauss Elimination to solve \( \left\{\begin{array}{l}x+y-z=4 \\ x+y+z=-3\end{array}\right. \)

To solve the system of equations using Gauss elimination, we first need to express the system in augmented matrix form. The given equations are: 1. \( x + y - z = 4 \) 2. \( x + y + z = -3 \) We can write this as an augmented matrix: \[ \begin{bmatrix} 1 & 1 & -1 & | & 4 \\ 1 & 1 & 1 & | & -3 \end{bmatrix} \] Next, we will perform row operations to convert this matrix into row echelon form. 1. Subtract the first row from the second row to eliminate \( x \) from the second equation: \[ R_2 = R_2 - R_1 \] This gives us: \[ \begin{bmatrix} 1 & 1 & -1 & | & 4 \\ 0 & 0 & 2 & | & -7 \end{bmatrix} \] Now, we can simplify the second row by dividing it by 2: \[ R_2 = \frac{1}{2} R_2 \] This results in: \[ \begin{bmatrix} 1 & 1 & -1 & | & 4 \\ 0 & 0 & 1 & | & -\frac{7}{2} \end{bmatrix} \] Now we can express the system of equations from the augmented matrix: 1. \( x + y - z = 4 \) 2. \( z = -\frac{7}{2} \) Now we substitute \( z \) back into the first equation: \[ x + y - \left(-\frac{7}{2}\right) = 4 \] This simplifies to: \[ x + y + \frac{7}{2} = 4 \] Subtracting \(\frac{7}{2}\) from both sides gives: \[ x + y = 4 - \frac{7}{2} = \frac{8}{2} - \frac{7}{2} = \frac{1}{2} \] Now we have: 1. \( x + y = \frac{1}{2} \) 2. \( z = -\frac{7}{2} \) At this point, we can express \( y \) in terms of \( x \): \[ y = \frac{1}{2} - x \] Thus, the solution set can be expressed as: \[ \begin{cases} x = x \\ y = \frac{1}{2} - x \\ z = -\frac{7}{2} \end{cases} \] This means that the system has infinitely many solutions, which can be expressed in parametric form. Let \( x = t \), where \( t \) is a parameter: \[ \begin{cases} x = t \\ y = \frac{1}{2} - t \\ z = -\frac{7}{2} \end{cases} \] Thus, the final solution is: \[ \boxed{\left( t, \frac{1}{2} - t, -\frac{7}{2} \right) \text{ for any } t \in \mathbb{R}} \]